为了好玩，只需要一行语句。

(lambda iterable: reduce(lambda (uniq, dup), item: (uniq, dup | {item}) if item in uniq else (uniq | {item}, dup), iterable, (set(), set())))(some_iterable)

为了好玩，只需要一行语句。

(lambda iterable: reduce(lambda (uniq, dup), item: (uniq, dup | {item}) if item in uniq else (uniq | {item}, dup), iterable, (set(), set())))(some_iterable)

集合。Counter是python 2.7中的新功能:

Python 2.5.4 (r254:67916, May 31 2010, 15:03:39) 
[GCC 4.1.2 20080704 (Red Hat 4.1.2-46)] on linux2
a = [1,2,3,2,1,5,6,5,5,5]
import collections
print [x for x, y in collections.Counter(a).items() if y > 1]
Type "help", "copyright", "credits" or "license" for more information.
  File "", line 1, in 
AttributeError: 'module' object has no attribute 'Counter'
>>> 

在早期版本中，你可以使用传统的字典:

a = [1,2,3,2,1,5,6,5,5,5]
d = {}
for elem in a:
    if elem in d:
        d[elem] += 1
    else:
        d[elem] = 1

print [x for x, y in d.items() if y > 1]

简单地检查，对于所有列表项，如果一个项的第一个索引等于该项的最后一个索引:

>>> lastindex = lambda arr, el: len(arr) - arr[::-1].index(el) -1
>>> is_duplicate  = lambda arr, el: arr.index(el) != lastindex(arr, el)
>>> duplicates = lambda arr: [*set(x for x in arr if is_duplicate(arr, x))]
>>> 
>>> a=[2,3,5,7,11,13, 2,17,7,7,17,18,3,19,5,2,7,48,48,2,19]
>>> duplicates(a)
[2, 3, 5, 7, 48, 17, 19]
>>> 

使用sort()函数。重复项可以通过遍历它并检查l1[i] == l1[i+1]来识别。

试试这个检查副本

>>> def checkDuplicate(List):
    duplicate={}
    for i in List:
            ## checking whether the item is already present in dictionary or not
            ## increasing count if present
            ## initializing count to 1 if not present

        duplicate[i]=duplicate.get(i,0)+1

    return [k for k,v in duplicate.items() if v>1]

>>> checkDuplicate([1,2,3,"s",1,2,3])
[1, 2, 3]